Question

(корень 3\2)sinx+(1\2)cosx=sin5x

Answer 1

$\frac{\sqrt{3} }{2}Sinx+\frac{1}{2}Cosx=Sin5x\\\\Cos\frac{\pi }{6}Sinx+Sin\frac{\pi }{6}Cosx=Sin5x\\\\Sin(x+\frac{\pi }{6})-Sin5x=0\\\\2Sin\frac{x+\frac{\pi }{6}-5x }{2}Cos\frac{x+\frac{\pi }{6}+5x }{2} =0\\\\Sin(\frac{\pi }{12} -2x)Cos(3x+\frac{\pi }{12})=0\\\\1)Sin(\frac{\pi }{12}-2x)=0\\\\-2x+\frac{\pi }{12}=\pi n,n\in z\\\\-2x=-\frac{\pi }{12}+\pi n,n\in z\\\\x=\frac{\pi }{24} -\frac{\pi n }{2},n\in z$

$2)Cos(3x+\frac{\pi }{12})=0\\\\3x+\frac{\pi }{12}=\frac{\pi }{2} +\pi n,n\in z\\\\3x=\frac{\pi }{2}-\frac{\pi }{12}+\pi n,n\in z\\\\3x=\frac{5\pi }{12}+\pi n,n\in z\\\\x=\frac{5\pi }{36}+\frac{\pi n }{3},n\in z$